# On convergence of certain integrals under weak convergence

As I work on certain problems that are relevant to me, I come across questions that make me realize how much just knowing the basic apparatus of results is not enough..

To simplify as much as possible. Let's say that we have a sequence of measures (not necessarily probability measures, but positive and one can even add the requirement that $\inf_n \mu_n(R)> 0$ if need be) $\{\mu_n: n \in N\}$ that are known (proved already) to converge weakly to a finite measure $\mu.$ Then we know that if $f$ is a bounded continuous function, $$\int f(x)\, \mu_n(dx) \to \int f(x)\, \mu(dx)$$ as $n \to \infty.$ What happens to terms like, say, $$\int_{|x|\le c} f(x)\, \mu_n(dx)$$ and/or $$\int_{|x| > c} f(x)\, \mu_n(dx)~?$$

Also, is the choice of the points $\pm c$ completely free or is it necessary to assume more structure like requiring that $c$ be a point of continuity for $\mu,$ i.e., $\mu(\{\pm c\}) = 0?$ In this case it would look like that if we consider the integral $$\int_{|x|\le c} f(x)\, \mu_n(dx)$$ then it would be possible to use the fact that $$\biggl| \int_{|x|\le c} f(x)\, \mu_n(dx) - \int_{|x|\le c} f(x)\, \mu(dx)\biggl| \le K |\mu_n([-c, c]) - \mu([-c,c])| \to 0$$ using the definition of weak convergence and the fact that $\mu(\partial [-c, c]) = 0$ where $\partial$ denotes the boundary of $[-c, c].$

And, if the requirement that $c$ be indeed a point of continuity for $\mu$ were necessary, is there any easy counter-example to the fact that we cannot choose the points $\pm c$ completely arbitrary?

If anyone has any wisdom to share on these issues, I would be deeply grateful.

Maurice

Consider e.g. $\mu_n(dx) = (\delta(x - 1 - 1/n) - \delta(x - 1 + 1/n))\; dx$ (i.e. a unit mass at $1+1/n$ minus a unit mass at $1-1/n$), which converges weakly to $\mu = 0$. But $$\int_{|x| > 1} f(x)\; \mu_n(dx) = \int_{|x| \ge 1} f(x)\; \mu_n(dx) = f(1 + 1/n) \to f(1)$$ So continuity of $\mu$ is not enough.
• Thank you Robert. In your counter-example, unless I am making a silly mistake, the measure you suggested would be such that $\mu_n(\{1-1/n\})=-1$ and hence $\mu_n$ would not be a measure. Dec 1, 2014 at 19:19